Question

a car moves with a speed of 50km/h. if stopping distance is 40m and its rate of retardation in 4.4m/s^2,then find the time of arrival​

Answer 1

Answer:

Here you go .............

Step-by-step explanation:

S = ut + 1/2 at²

s = 40m , a = -4.4m/s² , u = 50kmph = 50x5/18 m/s = 125/9 m/s

ON SUBSTITUTING THESE VALUES IN THE EQUATION WRITTEN WE GET t = 2.63 or t = -30.41

Since, time cannot be a negative quantity. So, t = -30.41 is rejected

Hence, t = 2.63 seconds

Therefore, the time of arrival = 2.63 seconds.

Hope it helps you.

Have a nice day.